A toy train rolls around a horizontal 1.0-m-diameter track. the coefficient of rolling friction is 0.10. how long does it take t

Question

Mnenie [13.5K]

3 years ago

5

A toy train rolls around a horizontal 1.0-m-diameter track. the coefficient of rolling friction is 0.10. how long does it take t

he train to stop if it's released with an angular speed of 40 rpm?

Physics

2 answers:

Musya8 [376] · Answer 1 · 2022-08-26T20:12:58+03:00

Musya8 [376]3 years ago

6 0

Deceleration due to rolling resistance (α) = coefficient*g/R = 0.1*9.81/0.5 = 1.962 rad/sec^2

Equations of motion:
ωf=ωo+αt
But ωf = 0 rad/sec; ωo= 2*pi*RPM/60 = 4.19 rad/sec

Therefore,
0=4.19-1.962t => 1.962t = 4.19 => t = 4.19/1.962 = 2.13 seconds

Zielflug [23.3K] · Answer 2 · 2022-08-26T21:28:31+03:00

The train with a constant angular deceleration will stop in $\fbox{2.13\text{ s}}$ .

Further Explanation:

An object in pure rolling motion on a rough surface experiences resistive rolling force.

Given:

The diameter of wheel is $1\,{\text{m}}$ .

The angular speed is $40\,{\text{rpm}}$ .

Concept:

The equation of resistive or friction force is:

$\fbox{\begin\\F_r=m\cdot\text{a}_r\end{minispace}}$

Here, $m$ is the mass of body, ${a_r}$ is the deceleration due to rolling resistance and ${F_r}$ is the frictional force.

The equation of normal force is:

$\fbox{N = mg}$

Here, $g$ is the gravitational acceleration, $m$ is the mass of body, and $N$ is the normal force (which is equal to weight of body).

For body to be in equilibrium the equation of forces is:

${F_r} = \dfrac{{bN}}{R}$

Here, $b$ is the coefficient of rolling friction and $R$ is the radius of wheel.

Substitute $m{a_r}$ for ${F_r}$ and $mg$ for $N$ in above equation.

$\begin{gathered}m{a_r}=\frac{{bmg}}{R}\\{a_r}=\frac{{bg}}{R}\\\end{gathered}$

Here, ${a_r}$ is the deceleration due to rolling resistance, $b$ is the coefficient of rolling friction, $g$ is the acceleration due to gravity and $R$ is the radius of wheel.

Substitute $0.1$ for $b$ , $9.81\text{ m}/\text{s}^2$ for $g$ , and $\dfrac{1}{2}$ for $R$ in above equation.

$\begin{gathered}{a_r}=\frac{{0.1\times9.81\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}}{{0.5}}\\=1.962\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{gathered}$

The initial angular velocity of body is:

${\omega _0} = \dfrac{{2\pi N}}{{60}}$

Here, $N$ is the angular speed in rpm and ${\omega _0}$ is the initial angular velocity.

Substitute $40\,{\text{rpm}}$ for $N$ in above equation.

$\begin{gathered}{\omega _0}=\frac{{2\pi\times40\,{\text{rpm}}}}{{60}}\\=4.19\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{gathered}$

Applying equation of motion:

${\omega _f}={\omega _0}-{\alpha _r}t$

Rearrange the above equation for value t:

$t=\dfrac{{{\omega _0}-{\omega _f}}}{{{\alpha _r}}}$

Here, ${\omega _f}$ is the final velocity, ${\omega _0}$ is the initial velocity, ${\alpha _r}$ is the deceleration and $t$ is the time taken by train to come to rest.

Substitute $0\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ for ${\omega _f}$ and $1.962\,{{{\text{rad}}} \mathord{\left/ {\vphantom {{{\text{rad}}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}$ for ${\alpha _r}$ and $4.19\,{{{\text{rad}}} \mathord{\left/ {\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${\omega _0}$ in above equation.

$\begin{aligned}\\t&=\frac{{4.19\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}-0}}{{1.962\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}}\\&=2.13\,{\text{s}}\\\end{aligned}$